The electrostatic energy of $$Z$$ protons uniformly distributed throughout a spherical nucleus of radius $$R$$ is given by
$$E = \frac{3}{5}\frac{{Z\left( {Z - 1} \right){e^2}}}{{4\pi {\varepsilon _0}R}}$$
The measured masses of the neutron $$_1^1H,\,_7^{15}N$$   and $$_8^{15}O$$ are $$1.008665\,u,\,1.007825\,u,\,15.000109\,u$$       and $$15.003065\,u,$$   respectively. Given that the radii of both the $$_7^{15}N$$ and $$_8^{15}O$$ nuclei are same, $$1\,u = 931.5\,Me\,V/{c^2}$$    ($$e$$ is the speed of light) and $$\frac{{{e^2}}}{{\left( {4\pi {\varepsilon _0}} \right)}} = 1.44\,MeV\,fm.$$     Assuming that the difference between the binding energies of $$_7^{15}N$$ and $$_8^{15}O$$  is purely due to the electrostatic energy, the radius of either of the nuclei is
$$\left( {1fm. = {{10}^{ - 15}}m} \right)$$

Solution :
Binding energy of nitrogen atom
$$ = \left[ {8 \times 1.008665 + 7 \times 1.007825 - 15.000109} \right] \times 931$$
Binding energy of oxygen atom
$$\eqalign{ & = \left[ {7 \times 1.008665 + 8 \times 1.007825 - 15.003065} \right] \times 931 \cr & \therefore {\text{Difference}} = 0.0037960 \times 931MeV\,......\left( {\text{i}} \right) \cr & {\text{Also}}\,{E_O} = \frac{3}{5} \times \frac{{8 \times 7}}{R} \times \frac{{{e^2}}}{{4\pi { \in _0}}} = \frac{3}{5} \times \frac{{56}}{R} \times 1.44MeV \cr & {E_N} = \frac{3}{5} \times \frac{{7 \times 6}}{R} \times \frac{{{e^2}}}{{4\pi { \in _0}}} = \frac{3}{5} \times \frac{{42}}{R} \times 1.44MeV \cr & \therefore {E_O} - {E_N} = \frac{3}{5} \times \frac{{14}}{R} \times 1.44MeV\,......\left( {{\text{ii}}} \right) \cr} $$
From (i) & (ii)
$$\eqalign{ & \frac{3}{5} \times \frac{{14}}{R} \times 1.44 = 0.0037960 \times 931 \cr & \therefore R = 3.42fm \cr} $$